# Natural Number Multiplication Distributes over Addition

## Theorem

The operation of multiplication is distributive over addition on the set of natural numbers $\N$:

$\forall x, y, z \in \N:$
$\paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
$z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

## Proof 1

 $\ds \paren {x + y} \times z$ $=$ $\ds +^z \paren {x + y}$ Definition of Natural Number Multiplication $\ds$ $=$ $\ds \paren {+^z x} + \paren {+^z y}$ Power of Product of Commuting Elements in Semigroup equals Product of Powers $\ds$ $=$ $\ds x \times z + y \times z$

$\Box$

 $\ds z \times \paren {x + y}$ $=$ $\ds +^{x + y} z$ Definition of Natural Number Multiplication $\ds$ $=$ $\ds \paren {+^x z} + \paren {+^y z}$ Index Laws for Semigroup: Sum of Indices $\ds$ $=$ $\ds z \times x + z \times y$

$\blacksquare$

## Proof 2

We are to show that:

$\forall x, y, z \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$

From the definition of natural number multiplication, we have by definition that:

 $\ds \forall m, n \in \N: \,$ $\ds m \times 0$ $=$ $\ds 0$ $\ds m \times n^+$ $=$ $\ds \paren {m \times n} + m$

Let $x, y \in \N$ be arbitrary.

For all $z \in \N$, let $\map P z$ be the proposition:

$\forall x, y \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$

### Basis for the Induction

$\map P 0$ is the case:

 $\ds \paren {x + y} \times 0$ $=$ $\ds 0$ Definition of Natural Number Multiplication‎ $\ds$ $=$ $\ds 0 + 0$ Definition of Natural Number Addition $\ds$ $=$ $\ds x \times 0 + y \times 0$ Definition of Natural Number Multiplication‎

and so $\map P 0$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.

So this is our induction hypothesis:

$\forall x, y \in \N: \paren {x + y} \times k = \paren {x \times k} + \paren {y \times k}$

Then we need to show:

$\forall x, y \in \N: \paren {x + y} \times k^+ = \paren {x \times k^+} + \paren {y \times k^+}$

### Induction Step

This is our induction step:

 $\ds \paren {x + y} \times k^+$ $=$ $\ds \paren {x + y} \times k + \paren {x + y}$ Definition of Natural Number Multiplication‎ $\ds$ $=$ $\ds \paren {x \times k} + \paren {y \times k} + \paren {x + y}$ Induction Hypothesis $\ds$ $=$ $\ds \paren {\paren {x \times k} + x} + \paren {\paren {y \times k} + y}$ Natural Number Addition is Commutative and Associative $\ds$ $=$ $\ds \paren {x \times k^+} + \paren {y \times k^+}$ Definition of Natural Number Multiplication

So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction:

$\forall x, y, z \in \N: \paren {x + y} \times n = \paren {x \times z} + \paren {y \times z}$

$\Box$

Next we need to show that:

$\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

So:

 $\ds z \times \paren {x + y}$ $=$ $\ds \paren {x + y} \times z$ Natural Number Multiplication is Commutative $\ds$ $=$ $\ds \paren {x \times z} + \paren {y \times z}$ from above $\ds$ $=$ $\ds \paren {z \times x} + \paren {z \times y}$ Natural Number Multiplication is Commutative

Thus we have proved:

$\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

$\blacksquare$

## Proof 3

In the Axiomatization of 1-Based Natural Numbers, this is rendered:

$\forall x, y, z \in \N_{> 0}:$
$\paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
$z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

Using the following axioms:

 $(\text A)$ $:$ $\ds \exists_1 1 \in \N_{> 0}:$ $\ds a \times 1 = a = 1 \times a$ $(\text B)$ $:$ $\ds \forall a, b \in \N_{> 0}:$ $\ds a \times \paren {b + 1} = \paren {a \times b} + a$ $(\text C)$ $:$ $\ds \forall a, b \in \N_{> 0}:$ $\ds a + \paren {b + 1} = \paren {a + b} + 1$ $(\text D)$ $:$ $\ds \forall a \in \N_{> 0}, a \ne 1:$ $\ds \exists_1 b \in \N_{> 0}: a = b + 1$ $(\text E)$ $:$ $\ds \forall a, b \in \N_{> 0}:$ $\ds$Exactly one of these three holds: $\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y}$ $(\text F)$ $:$ $\ds \forall A \subseteq \N_{> 0}:$ $\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0}$

### Left Distributive Law for Natural Numbers

First we show that:

$n \times \paren {x + y} = \paren {n \times x} + \paren {n \times y}$

Let us cast the proposition in the form:

$\forall a, b, n \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\forall a, b \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$

### Basis for the Induction

$\map P 1$ is the case:

 $\ds a \times \paren {b + 1}$ $=$ $\ds \paren {a \times b} + a$ Axiom $\text B$ $\ds$ $=$ $\ds \paren {a \times b} + \paren {a \times 1}$ Axiom $\text A$

and so $\map P 1$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\forall a, b \in \N_{> 0}: a \times \paren {b + k} = \paren {a \times b} + \paren {a \times k}$

Then we need to show:

$\forall a, b \in \N_{> 0}: a \times \paren {b + \paren {k + 1} } = \paren {a \times b} + \paren {a \times \paren {k + 1} }$

### Induction Step

This is our induction step:

 $\ds a \times \paren {b + \paren {k + 1} }$ $=$ $\ds a \times \paren {\paren {b + k} + 1}$ Axiom $\text C$ $\ds$ $=$ $\ds \paren {a \times \paren {b + k} } + a$ Axiom $\text B$ $\ds$ $=$ $\ds \paren {\paren {a \times b} + \paren {a \times k} } + a$ Induction hypothesis $\ds$ $=$ $\ds \paren {a \times b} + \paren {\paren {a \times k} + a}$ Natural Number Addition is Associative $\ds$ $=$ $\ds \paren {a \times b} + \paren {a \times \paren {k + 1} }$ Axiom $\text B$

The result follows by the Principle of Mathematical Induction.

$\Box$

### Right Distributive Law for Natural Numbers

Then we show that:

$\paren {x + y} \times n = \paren {x \times n} + \paren {y \times n}$

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times n = \paren {a \times n} + \paren {b \times n}$

### Basis for the Induction

$\map P 1$ is the case:

 $\ds \paren {a + b} \times 1$ $=$ $\ds a + b$ Axiom $\text A$ $\ds$ $=$ $\ds \paren {a \times 1} + \paren {b \times 1}$ Axiom $\text A$

and so $\map P 1$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times k = \paren {a \times k} + \paren {b \times k}$

Then we need to show:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times \paren {k + 1} = \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }$

### Induction Step

This is our induction step:

 $\ds \paren {a + b} \times \paren {k + 1}$ $=$ $\ds \paren {\paren {a + b} \times k} + \paren {a + b}$ Axiom $\text B$ $\ds$ $=$ $\ds \paren {\paren {a \times k} + \paren {b \times k} } + \paren {a + b}$ Induction hypothesis $\ds$ $=$ $\ds \paren {\paren {a \times k} + a} + \paren {\paren {b \times k} + b}$ Natural Number Addition is Commutative $\ds$ $=$ $\ds \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }$ Axiom $\text B$

The result follows by the Principle of Mathematical Induction.

$\Box$

The result follows.

$\blacksquare$